Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON3):

Solution:
N = int(input())
p = list(map(lambda x: float(x), input().split()))

#dp[i][h] = i+1番目までで表h枚の確率
dp = [[0]*(N+1) for i in range(N)]
dp[0][0] = 1-p[0]
dp[0][1] = p[0]

for i in range(1, N):
  for h in range(N+1):
    dp[i][h] = dp[i-1][h]*(1-p[i]) + (0 if h==0 else dp[i-1][h-1]*p[i])
    
print(sum(dp[N-1][int((N+1)/2):]))